I have this function
$f(x,y)=3y^2-12x^2+1$ with the constraint $h(x,y)=x^2+y^2-x-2y+\frac54=0$.
First thing I do is write
$$3y^2-12x^2+1-λ(x^2+y^2-x-2y+\frac54)$$ which expands to $$3y^2-12x^2+1-λx^2-λy^2+λx+2yλ-\frac54.$$
The partial derivative with respect to $x$:
$-24x-2λx+λ=0$ where I get that $x= \dfrac λ{2(12+λ)}$.
The partial derivative with respect to $y$:
$6y-2yλ+2λ=0$ where I get that $y= -\dfrac λ{3-λ}$.
The partial derivative with respect to $λ$:
$$-x^2-y^2+x+2y$$
Now I know that I need to plug in the values of $x$ and $y$ in the $λ$ partial derivative which is $-x^2-y^2+x+2y$.
And here is where my problems arise.
On the answer sheet it states that $y= \dfrac λ{3-λ}$ (different from my answer since I get a negative $λ$ on the numerator).
Consequently, it states that I need to plug in the values of $x$ and $y$ NOT in the partial derivative of $λ$ (which I thought was always the case) but it states that I need to plug the values of $x$ and $y$ in the CONSTRAINT, hence $x^2+y^2-x-2y+\frac54$.
So my two questions are :
1) Why am I getting a different $y$ value?
2) Who is in the wrong here? Do I need to plug the $x$ and $y$ values in the partial derivative of $λ$ or in the constraint formula? Why?
Thanks in advance for the help. I have some learning disabilities involving maths so I would really appreciate your answers being as simple and clear as possible. Thanks!!
The $\lambda$ partial derivative should have the $5/4$ still in it. The constraint and $dL/d\lambda$ should be the same.
I usually add $\lambda$ times the constraint instead of subtracting it so my $\lambda$ would be the negative of yours.